Here s the problem:
Obtain an expression for the distance s travelled by a body travelling with a velocity given by v(t)=t^2 -4t +4 m/s
And establish the distance travelled in the first 2 seconds
Thanks
How can I solve this integral calculus problem?
2016-06-21 7:47 am
回答 (2)
2016-06-21 7:59 am
Distance travelled, s(t)
= ∫ v(t) dt
= ∫ (t² - 4t + 4) dt m
= [(1/3)t³ - 2t² + 4t + C] m
Distance travelled in the first two seconds
= ∫(0→2) v(t) dt
= [(1/3)(2)³ - 2(2)² + 4(2)] - [(1/3)(0)³ - 2(0)² + 4(0)] m
= 8/3 m
= ∫ v(t) dt
= ∫ (t² - 4t + 4) dt m
= [(1/3)t³ - 2t² + 4t + C] m
Distance travelled in the first two seconds
= ∫(0→2) v(t) dt
= [(1/3)(2)³ - 2(2)² + 4(2)] - [(1/3)(0)³ - 2(0)² + 4(0)] m
= 8/3 m
2016-06-21 7:55 am
It's easy. V= ds/dt , multiply the right side with dt and integrate with respect to t to get the value of s. Then put t=2 , you would get your distance. Good luck
收錄日期: 2021-04-18 15:09:10
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